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The question:

$a,b,c > 0; ab+bc+ca =3$, Prove that

$\sum_{cyc}\frac{a}{\sqrt{a^{3}+5}}\leq\frac{\sqrt{6}}{2}$

The sum is cyclic over $a,b,c$

I've looked at the problem for a long time but still can't think of an approach for this, so how can I solve this?

The Social Pi
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1 Answers1

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We have \begin{align*} \sum_{\mathrm{cyc}} \frac{a}{\sqrt{a^3 + 5}} &= \sum_{\mathrm{cyc}} \frac{a}{\sqrt{\frac{a^3}{2} + \frac{a^3}{2} + \frac12 + \frac92}}\\ &\le \sum_{\mathrm{cyc}} \frac{a}{\sqrt{3\sqrt[3]{\frac{a^3}{2} \cdot \frac{a^3}{2} \cdot \frac12} + \frac92}}\\ &= \sum_{\mathrm{cyc}} \frac{2a}{\sqrt{6a^2 + 18}}\\ &= \sum_{\mathrm{cyc}} \frac{2a}{\sqrt{6a^2 + 6(ab + bc + ca)}}\\ &= \frac{1}{\sqrt{6}}\sum_{\mathrm{cyc}} 2\sqrt{\frac{a}{a + b}} \sqrt{\frac{a}{a + c}}\\ &\le \frac{1}{\sqrt{6}}\sum_{\mathrm{cyc}} \left(\frac{a}{a + b} + \frac{a}{a + c}\right)\\ &= \frac{\sqrt{6}}{2}. \end{align*}

We are done.

River Li
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  • How did you think to come up with that? (I agree that the steps are valid.) – Calvin Lin Dec 06 '21 at 22:30
  • @CalvinLin First, since the constraint is degree 2, the term $a^3 + 5$ is not good to deal with (e.g. homogenization), so I want to make it become degree 2. This was done by $a^3\ge \frac{3a^2 - 1}{2}$. Then I found that $6a^2 + 18 = 6a^2 + 6(ab + bc + ca)$ (homogenization) is further factorized to $6(a + b)(a + c)$. – River Li Dec 07 '21 at 01:35
  • An awesome answer IMHO @RiverLi: AGM only is used, twice & in a flip-flopping way, first in the denominator, then forward-oriented towards the final stroke. I wonder if there's a similarly quick one, based on Cauchy–Bunyakovsky–Schwarz ?! – Hanno Dec 07 '21 at 08:04
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    @Hanno Thanks. I just now used Approach Zero to search it. It is in JBMO 2018 shortlist: https://artofproblemsolving.com/community/c6h1861682p12594883 – River Li Dec 07 '21 at 11:49